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Please help me to prove the inequality $$ \sqrt{a^2 + b^2} \geq \frac{|a-b|}{\sqrt{2}}. $$

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Hint: Square both sides and multiply by $2$, and you'll find that this is equivalent to proving that

$$2(a^2 + b^2) \ge |a - b|^2 = (a - b)^2$$

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Hint: Note that $(a-b)^2+(a+b)^2=2(a^2+b^2)$.

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